Proof of a conjectured formula for A208531
from Toshitaka Suzuki (suzuki(AT)scio.co.jp)
July 29 2019

Theorem: A208531(n) = (2*n^3+3*n^2+n-366)/6 for n > 8.

Proof:
Define:
issum(m, n) =
1 if m is the sum of distinct squares of 1,4,9,…,n^2,
0 otherwise.

By computing all combinations,
issum(k, 9) = 1 for 0 <= k <= 128 except A001422.
And computing all combinations,
issum(k, 10) = 1 for 129 <= k <= 256.
Then issum(k, 11) = 1 for 129+121 <= k <= 256+121
by adding 11^2=121,
so issum(k, 11) = 1 for 129 <= k <= 377.
Similarly,
issum(k, 12) = 1 for 129+144 <= k <= 377+144
so issum(k, 12) = 1 for 129 <= k <= 521
issum(k, 13) = 1 for 129+169 <= k <= 521+169
so issum(k, 13) = 1 for 129 <= k <= 690
...
issum(k, n) = 1 for 129 <= k <= 256 + Sum_{m=11..n}m^2

# Because
# (256+Sum_{m=11..n}m^2)-129+1 - (n+1)^2
# = (2(n-10)^3+57(n-10)^2+529(n-10)+42)/6 > 0

256 + Sum_{m=11..n}m^2 > (Sum_{m=1..n}m^2) / 2 for n >= 10 and
issum(m, n) = issum((Sum_{m=1..n}m^2) - m, n) (by reverse use or not)
so
issum(k, n) = 1 for 0 <= k <= Sum_{m=1..n}m^2
except A001422 and (Sum_{m=1..n}m^2) - A001422 for n >= 10.
Therefore
A208531(n) = (Sum_{m=1..n}m^2) + 1 - 31*2 = (2*n^3+3*n^2+n-366)/6
for n >= 10 (31 is not in A208531 )
and A208531(9)=(2*9^3+3*9^2+9-366)/6,
hence A208531(n) = (2*n^3+3*n^2+n-366)/6 for n > 8.
QED